The points that are 5 units away from the point (2, 7) form a circle, since the definition of a circle is the set of points so many units away from a point.
Remembering our circle formula, the equation of the circle is:
\({(x-2)}^{2}+{(y-7)}^{2}=25\).
Any point that is on the circle and \(y=5x-28\), must satisfy both equations, meaning it is the solution of the system:
\(\begin{cases} {(x-2)}^{2}+{(y-7)}^{2}=25 \\ y=5x-28 \end{cases}\).
Substitute:
\({(x-2)}^{2}+{(5x-35)}^{2}=25\)
\({(x-2)}^{2}+25{(x-7)}^{2}=25\)
\(26{x}^{2}-354x+1204=0\)
\((13x-86)(x-7)=0\)
\(x_{1, 2}= 7, \frac{86}{13}\)
Plugging back in,
\(\begin{cases} x=7, y=7\\x=\frac{86}{13},y=\frac{66}{13}\end{cases}\)
Our two points are \(\bf{(7, 7), (\frac{86}{13}, \frac{66}{13})}\).